Estude! Com Prof. Caju

Num triângulo ABC, a altura AH forma com a bissetriz interna AS um ângulo de 20° e a bissetriz dos angulos externos B e C formam um angulo de 30°. calcular o maior angulo do triangulo.

a)130°
b)120°
c)150°
d)144°
e)110°

[tex3]\angle H=90^{\circ}[/tex3]
[tex3]\angle S=180-90-20=70^{\circ}[/tex3]
[tex3]\angle ASB=110^{\circ}[/tex3]

[tex3]\angle BAS= \angle HAC =\theta[/tex3]
[tex3]\angle ABS=180^{\circ}-2\beta[/tex3]
[tex3]\angle ACH=180^{\circ}-2\alpha[/tex3]

No triângulo BCM,
[tex3]\alpha+\beta=150^{\circ}[/tex3]

No triângulo ABC,
[tex3]2\theta+180^{\circ}-2\alpha+180^{\circ}-2\beta=180^{\circ}[/tex3]
[tex3]2\theta-2\alpha-2\beta = -180^{\circ}[/tex3] [tex3]\div (-2)[/tex3]
[tex3]\underbrace{\alpha+\beta}_{150^{\circ}}-\theta=90^{\circ}[/tex3]
[tex3]\theta=150^{\circ}-90^{\circ}=60^{\circ}[/tex3]

Assim,
[tex3]\angle BAC=2\theta=120^{\circ}[/tex3]
[tex3]\angle ABS=180^{\circ}-110^{\circ}-60^{\circ}=10^{\circ}[/tex3]
[tex3]\angle ACB=180^{\circ}-60^{\circ}-70^{\circ}=50^{\circ}[/tex3]

Letra B